Unit 1 Exam Practice
Exponents and Logarithms
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
SL questions
1 Paper 1
Consider the equation \(3^{2x + 1} - 28\left(3^{x}\right) + 9 = 0\).
(a) Show that the equation can be written as \(3u^{2} - 28u + 9 = 0\), where \(u = 3^{x}\). [2]
(b) Hence solve the equation. [4]
Total 6 marks
(a)
\(3^{2x + 1} = 3 \times 3^{2x}\) M1
\(= 3\left(3^{x}\right)^{2} = 3u^{2}\), so \(3u^{2} - 28u + 9 = 0\) A1 AG
[2 marks]
(b)
\((3u - 1)(u - 9) = 0\) M1
\(u = \dfrac{1}{3}\) or \(u = 9\) A1
\(3^{x} = 3^{-1}\) or \(3^{x} = 3^{2}\) M1
\(x = -1\) or \(x = 2\) A1
[4 marks]
See 1.1.
2 Paper 1
Let \(\log_a x = p\), where \(a > 1\) and \(x > 0\).
(a) Given also that \(\log_a y = q\), express \(\log_a \left(\dfrac{x^{3}}{\sqrt{y}}\right)\) in terms of \(p\) and \(q\). [2]
(b) Express \(\log_{a^{2}} (ax)\) in terms of \(p\). [3]
(c) Given that \(\log_a x = 2\log_x a\) and \(x > a\), find the exact value of \(p\). Hence write \(x\) in terms of \(a\). [3]
Total 8 marks
(a)
\(3\log_a x - \dfrac{1}{2}\log_a y\) M1
\(= 3p - \dfrac{q}{2}\) A1
[2 marks]
(b)
\(\log_{a^{2}} (ax) = \dfrac{\log_a (ax)}{\log_a a^{2}}\) M1
\(= \dfrac{\log_a a + \log_a x}{2}\) A1
\(= \dfrac{1 + p}{2}\) A1
[3 marks]
(c)
\(\log_x a = \dfrac{1}{\log_a x} = \dfrac{1}{p}\), so \(p = \dfrac{2}{p}\) M1
\(p^{2} = 2\), and \(x > a > 1\) means \(p > 1\), so \(p = \sqrt{2}\) A1
\(x = a^{\sqrt{2}}\) A1
Rejecting \(p = -\sqrt{2}\) needs a reason.
[3 marks]
See 1.4.
3 Paper 1
Solve the equation \(\log_3 (x + 5) + 2\log_9 (x - 1) = 1 + \log_3 (2x - 1)\), giving your answer in the form \(a + \sqrt{b}\), where \(a, b \in \mathbb{Z}\). [7]
Total 7 marks
\(2\log_9 (x - 1) = \dfrac{2\log_3 (x - 1)}{\log_3 9} = \log_3 (x - 1)\) M1 A1
\(\log_3 \big((x + 5)(x - 1)\big) = \log_3 3 + \log_3 (2x - 1) = \log_3 \big(3(2x - 1)\big)\) M1
\((x + 5)(x - 1) = 3(2x - 1)\) A1
\(x^{2} + 4x - 5 = 6x - 3 \;\Rightarrow\; x^{2} - 2x - 2 = 0\) A1
\(x = 1 \pm \sqrt{3}\) A1
\(x\) must be greater than \(1\) (so that \(\log_3 (x - 1)\) is defined), so \(x = 1 + \sqrt{3}\) R1
Do not award the final R1 if \(x = 1 - \sqrt{3}\) is also given as an answer.
[7 marks]
See 1.5.
4 Paper 2
A cup of tea is left to cool in a room. Its temperature, \(T\) °C, \(t\) minutes after it is poured, is modelled by \[ T(t) = 20 + 70\,\mathrm{e}^{-kt}, \qquad t \geq 0, \] where \(k\) is a positive constant.
(a) Write down the temperature of the tea when it is poured. [1]
After \(5\) minutes the temperature of the tea is \(62\) °C.
(b) Show that \(k = \dfrac{1}{5}\ln\left(\dfrac{5}{3}\right)\). [3]
(c) Find the time taken for the tea to cool to \(40\) °C. [3]
(d) Write down the equation of the horizontal asymptote of the graph of \(T\), and explain what it means in this context. [2]
A cup of coffee is poured at the same time. Its temperature, \(C\) °C, is modelled by \(C(t) = 20 + 50\,(0.97)^{t}\).
(e) Find the time at which the tea and the coffee have the same temperature, and find this temperature. [3]
Total 12 marks
(a)
\(T(0) = 90\) °C A1
[1 mark]
(b)
\(62 = 20 + 70\,\mathrm{e}^{-5k}\) M1
\(\mathrm{e}^{-5k} = \dfrac{42}{70} = \dfrac{3}{5}\) A1
\(-5k = \ln\left(\dfrac{3}{5}\right)\), so \(k = -\dfrac{1}{5}\ln\left(\dfrac{3}{5}\right) = \dfrac{1}{5}\ln\left(\dfrac{5}{3}\right)\) A1 AG
[3 marks]
(c)
\(40 = 20 + 70\,\mathrm{e}^{-kt}\) M1
\(\mathrm{e}^{-kt} = \dfrac{2}{7}\), so \(t = \dfrac{\ln 3.5}{k}\) A1
\(t = 12.3\) minutes \((12.2621\ldots)\) A1
[3 marks]
(d)
\(T = 20\) A1
the temperature of the tea gets closer and closer to \(20\) °C (room temperature) but never reaches it R1
[2 marks]
(e)
\(20 + 70\,\mathrm{e}^{-kt} = 20 + 50\,(0.97)^{t}\) (for example, the intersection of the two graphs on a GDC) M1
\(t = 4.69\) minutes \((4.6923\ldots)\) A1
temperature \(= 63.3\) °C A1
[3 marks]
See 1.2.
5 Paper 1
Let \(f(x) = 3 - \ln(2x - 4)\), for \(x > 2\).
(a) Find the exact \(x\)-intercept of the graph of \(f\). [2]
(b) Write down the equation of the vertical asymptote of the graph of \(f\). [1]
(c) Find \(f^{-1}(x)\), and state the domain and the range of \(f^{-1}\). [5]
(d) Show that \(f(x) = c - \ln(x - 2)\), where \(c\) is a constant to be found in exact form. Hence describe a sequence of transformations that maps the graph of \(y = \ln x\) onto the graph of \(y = f(x)\). [4]
(e) Solve the equation \(f(x) = \ln x\), giving your answer in the form \(1 + \sqrt{1 + \dfrac{\mathrm{e}^{m}}{n}}\), where \(m, n \in \mathbb{Z}^{+}\). [4]
Total 16 marks
(a)
\(\ln(2x - 4) = 3 \;\Rightarrow\; 2x - 4 = \mathrm{e}^{3}\) M1
\(x = \dfrac{\mathrm{e}^{3} + 4}{2}\) A1
[2 marks]
(b)
\(x = 2\) A1
[1 mark]
(c)
\(x = 3 - \ln(2y - 4)\) (interchanging \(x\) and \(y\)) M1
\(2y - 4 = \mathrm{e}^{3 - x}\) A1
\(f^{-1}(x) = \dfrac{\mathrm{e}^{3 - x} + 4}{2}\) A1
domain of \(f^{-1}\): \(x \in \mathbb{R}\) A1
range of \(f^{-1}\): \(f^{-1}(x) > 2\) A1
[5 marks]
(d)
\(\ln(2x - 4) = \ln 2 + \ln(x - 2)\) M1
\(f(x) = (3 - \ln 2) - \ln(x - 2)\), so \(c = 3 - \ln 2\) A1
reflection in the \(x\)-axis A1
then a translation by the vector \(\begin{pmatrix} 2 \\ 3 - \ln 2 \end{pmatrix}\) A1
Accept the translation first if it is by \(\begin{pmatrix} 2 \\ \ln 2 - 3 \end{pmatrix}\), followed by the reflection.
[4 marks]
(e)
\(3 - \ln(2x - 4) = \ln x \;\Rightarrow\; \ln\big(x(2x - 4)\big) = 3\) M1
\(2x^{2} - 4x - \mathrm{e}^{3} = 0\) A1
\(x = \dfrac{4 \pm \sqrt{16 + 8\mathrm{e}^{3}}}{4} = 1 \pm \sqrt{1 + \dfrac{\mathrm{e}^{3}}{2}}\) M1
\(x > 2\), so \(x = 1 + \sqrt{1 + \dfrac{\mathrm{e}^{3}}{2}}\) A1
[4 marks]
See 1.5.
HL questions
HL questions on this unit test the same syllabus content as SL, but they need longer chains of reasoning.
1 Paper 1
Solve the equation \(2^{x} \times 3^{2x + 1} = 6^{x + 2}\), giving your answer in the form \(\log_3 a\), where \(a \in \mathbb{Z}^{+}\). [5]
Total 5 marks
taking \(\ln\) (or any base) of both sides: \(x\ln 2 + (2x + 1)\ln 3 = (x + 2)(\ln 2 + \ln 3)\) M1 A1
\(x\ln 2 + 2x\ln 3 + \ln 3 = x\ln 2 + x\ln 3 + 2\ln 2 + 2\ln 3\)
\(x\ln 3 = 2\ln 2 + \ln 3\) A1
\(x = \dfrac{\ln 4 + \ln 3}{\ln 3} = \dfrac{\ln 12}{\ln 3}\) M1
\(x = \log_3 12\) A1
[5 marks]
See 1.4.
2 Paper 1
Solve the simultaneous equations \[ \log_x y + \log_y x = \frac{5}{2}, \qquad xy = 27, \] where \(x > 1\) and \(y > 1\). [7]
Total 7 marks
let \(u = \log_x y\); then \(\log_y x = \dfrac{1}{u}\) M1
\(u + \dfrac{1}{u} = \dfrac{5}{2} \;\Rightarrow\; 2u^{2} - 5u + 2 = 0\) A1
\((2u - 1)(u - 2) = 0\), so \(u = 2\) or \(u = \dfrac{1}{2}\) A1
\(u = 2\): \(y = x^{2}\), so \(x^{3} = 27\) M1
\(x = 3,\ y = 9\) A1
\(u = \dfrac{1}{2}\): \(y = \sqrt{x}\), so \(x^{\frac{3}{2}} = 27\) M1
\(x = 9,\ y = 3\) A1
[7 marks]
See 1.4.
3 Paper 1
Let \(a = \log_2 3\) and \(b = \log_3 5\).
(a) Show that \(\log_{12} 18 = \dfrac{1 + 2a}{2 + a}\). [4]
(b) Express \(\log_{12} 30\) in terms of \(a\) and \(b\). [3]
Total 7 marks
(a)
\(\log_{12} 18 = \dfrac{\log_2 18}{\log_2 12}\) M1
\(\log_2 18 = \log_2 (2 \times 3^{2}) = 1 + 2a\) A1
\(\log_2 12 = \log_2 (2^{2} \times 3) = 2 + a\) A1
so \(\log_{12} 18 = \dfrac{1 + 2a}{2 + a}\) A1 AG
[4 marks]
(b)
\(\log_2 5 = \log_2 3 \times \log_3 5 = ab\) M1
\(\log_2 30 = \log_2 (2 \times 3 \times 5) = 1 + a + ab\) A1
\(\log_{12} 30 = \dfrac{1 + a + ab}{2 + a}\) A1
[3 marks]
See 1.4.
4 Paper 1
Consider the equation \(\mathrm{e}^{2x} - (k + 2)\,\mathrm{e}^{x} + 2k = 0\), where \(k \in \mathbb{R}\).
(a) Show that the equation can be written as \(\left(\mathrm{e}^{x} - 2\right)\left(\mathrm{e}^{x} - k\right) = 0\). [2]
(b) Solve the equation when \(k = 5\). [2]
(c) Find the set of values of \(k\) for which the equation has exactly two distinct real solutions. [3]
(d) Find the values of \(k\) for which one solution of the equation is twice the other. [4]
Total 11 marks
(a)
\(\left(\mathrm{e}^{x} - 2\right)\left(\mathrm{e}^{x} - k\right) = \mathrm{e}^{2x} - k\,\mathrm{e}^{x} - 2\,\mathrm{e}^{x} + 2k\) M1
\(= \mathrm{e}^{2x} - (k + 2)\,\mathrm{e}^{x} + 2k\) A1 AG
[2 marks]
(b)
\(\mathrm{e}^{x} = 2\) or \(\mathrm{e}^{x} = 5\) M1
\(x = \ln 2\) or \(x = \ln 5\) A1
[2 marks]
(c)
\(\mathrm{e}^{x} = 2\) always gives \(x = \ln 2\); \(\mathrm{e}^{x} = k\) has a solution only if \(k > 0\) R1
the two solutions are the same if \(k = 2\) R1
\(k > 0,\ k \neq 2\) A1
[3 marks]
(d)
the solutions are \(\ln 2\) and \(\ln k\) M1
\(\ln k = 2\ln 2 \;\Rightarrow\; k = 4\) A1
\(\ln 2 = 2\ln k \;\Rightarrow\; k^{2} = 2\), and \(k > 0\) M1
\(k = \sqrt{2}\) A1
[4 marks]
5 Paper 2
The number of bacteria in culture A, \(t\) hours after an experiment starts, is modelled by \(N = 2500\,\mathrm{e}^{0.42t}\).
(a) Find the number of bacteria in culture A after \(10\) hours. Give your answer in the form \(a \times 10^{k}\), where \(1 \leq a < 10\) and \(k \in \mathbb{Z}\). [2]
(b) Find the time taken for culture A to reach \(1.0 \times 10^{7}\) bacteria. [2]
The number of bacteria in culture B is modelled by \(M = 8.0 \times 10^{5} \times (1.15)^{t}\).
(c) Find the value of \(t\) at which the two cultures have the same number of bacteria. [3]
(d) Show that the graph of \(\ln N\) against \(t\) is a straight line. Write down its gradient and its vertical intercept. [3]
For culture C, the graph of \(\log_{10} P\) against \(t\) is a straight line through the points \((0,\ 3.2)\) and \((8,\ 5.6)\), where \(P\) is the number of bacteria.
(e) Find \(P\) in the form \(P = A \times b^{t}\), giving \(A\) and \(b\) correct to 3 s.f. [4]
Total 14 marks
(a)
\(N = 2500\,\mathrm{e}^{4.2} = 166\,715.8\ldots\) M1
\(1.67 \times 10^{5}\) A1
[2 marks]
(b)
\(2500\,\mathrm{e}^{0.42t} = 10^{7}\), so \(t = \dfrac{\ln 4000}{0.42}\) M1
\(t = 19.7\) hours \((19.747\ldots)\) A1
[2 marks]
(c)
\(2500\,\mathrm{e}^{0.42t} = 8.0 \times 10^{5} \times (1.15)^{t}\) M1
\(0.42t - t\ln 1.15 = \ln 320\) (or use the intersection on a GDC) A1
\(t = 20.6\) \((20.583\ldots)\) A1
[3 marks]
(d)
\(\ln N = \ln 2500 + \ln \mathrm{e}^{0.42t} = \ln 2500 + 0.42t\) M1
this has the form \(Y = mt + c\), so the graph is a straight line R1
gradient \(0.42\), vertical intercept \(\ln 2500\) \((7.82)\) A1
[3 marks]
(e)
gradient \(= \dfrac{5.6 - 3.2}{8} = 0.3\), so \(\log_{10} P = 3.2 + 0.3t\) M1 A1
\(P = 10^{3.2} \times \left(10^{0.3}\right)^{t}\) M1
\(P = 1580 \times 2.00^{t}\) \((A = 1584.89\ldots,\ b = 1.9952\ldots)\) A1
[4 marks]